（扰动法）
[egin{aligned}
&S(k,n)\
=&sum_{i=1}^ni^ka^i\
=&sum_{i=1}^n(i+1)^ka^{i+1}-(n+1)^ka^{n+1}+a\
=&sum_{i=1}^nsum_{j=0}^k{kchoose j}i^ja^{i+1}-(n+1)^ka^{n+1}+a\
=&asum_{j=0}^k{kchoose j}sum_{i=1}^ni^ja^i-(n+1)^ka^{n+1}+a\
=&asum_{j=0}^k{kchoose j}S(j,n)-(n+1)^ka^{n+1}+a
end{aligned}]
得到
[(1-a)S(k,n)=asum_{j=0}^{k-1}{kchoose j}S(j,n)-(n+1)^ka^{n+1}+a
]
[S(k,n)=frac{(n+1)^ka^{n+1}-asum_{j=0}^{k-1}{kchoose j}S(j,n)-a}{a-1}
]
但是必须要 (a e 1)。如果 (a=1) 呢？发现是 (k) 次幂和。
注意到推导中
[sum_{j=0}^{k-1}{kchoose j}S(j,n)=(n+1)^k-1
]
[{kchoose k-1}S(k-1,n)=(n+1)^k-sum_{j=0}^{k-2}{kchoose j}S(j,n)-1
]
最后一步换元，把系数除过去
[S(k,n)=frac{(n+1)^{k+1}-sum_{j=0}^{k-1}{k+1choose j}S(j,n)-1}{k+1}
]
注意边界条件 (S(0,n)=n)